# Difference between a spanning and linear combination?

1. Aug 15, 2010

### ichigo444

what is the difference of a span of a vector and a linear combination of a vector?

2. Aug 15, 2010

### quasar987

Given a set of vectors v_1,...,v_n, a linear combination of those vectors is a vector of the form a_1v_1+...+a_nv_n, for a_1,...,a_n some scalars. The span of the vectors v_1,...,v_n is the set of all the linear combinations of those vectors. I.e., it is the set whose elements are the vectors of the form a_1v_1+...+a_nv_n for all possible values of the coefficients a_1,...,a_n.

3. Aug 15, 2010

### annoymage

hmm, sorry for interrupting, but how to know when the span of some vector are equal?
ie: The span of the vectors v_1,...,v_n = The span of the vectors u_1,...,u_k.

4. Aug 15, 2010

### quasar987

One way to do it would be to check that every vector u_i is itself a linear combination of the v_j, and conversely, that every vector v_j can be written as a linear combination of the u_i.

If this is so, then span(u_1,...,u_k)=span(v_1,...,v_n). If not, then the spans are not equal.

Make sure you see why.

5. Aug 15, 2010

### annoymage

i can't clearly see it, every element in span(u_1,...,u_k) is then the element of span(v_1,...,v_n), and conversely,

then span(u_1,...,u_k) is subset of span(v_1,...,v_n), and conversely

is that really correct?

6. Aug 15, 2010

### quasar987

Suppose that for each i=1,...,k,

$$u_i = \sum_{j=1}^nc_i^jv_j$$

Then, for an arbitrary linear combination of the u_i,

$$\sum_{i=1}^ka^iu_i=\sum_{i=1}^ka^i\left(\sum_{j=1}^nc_i^jv_j\right)=\sum_{j=1}^n\left(\sum_{i=1}^ka^ic_i^j\right)v_j$$

(a linear combination of the v_j !) This shows that $\mathrm{span}(u_1,\ldots,u_k)\subset \mathrm{span}(v_1,\ldots,v_n)$.

And in the same way, if each v_j can be written as a linear combination of the u_i, we obtain $\mathrm{span}(v_1,\ldots,v_n)\subset \mathrm{span}(u_1,\ldots,u_k)$.

And so in that case, $\mathrm{span}(v_1,\ldots,v_n)= \mathrm{span}(u_1,\ldots,u_k)$.

On the other hand, if for instance, u_i cannot be written as a linear combination of the v_j's, then $\mathrm{span}(v_1,\ldots,v_n)\neq \mathrm{span}(u_1,\ldots,u_k)$ since $u_i\in \mathrm{span}(u_1,\ldots,u_k)$ but $u_i \notin\mathrm{span}(v_1,\ldots,v_n)$.

7. Aug 15, 2010

### annoymage

thanksssssssssss, i can see it now

8. Aug 15, 2010

### HallsofIvy

Staff Emeritus
And please note that these are the span of a set of vectors and the linear combination of a set of vectors. Typically the set contains more than just one vector!

A linear combination is single sum of scalars times vectors in the set. The span is the collection of all possible linear combinations of the set.